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Commit 83c4a31

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Merge branch 'master' of github.com:youngyangyang04/leetcode-master
2 parents a64f7a1 + fa66994 commit 83c4a31

22 files changed

+445
-189
lines changed

‎problems/0028.实现strStr.md‎

Lines changed: 2 additions & 2 deletions
Original file line numberDiff line numberDiff line change
@@ -9,7 +9,7 @@
99
1010
# 28. 实现 strStr()
1111

12-
[力扣题目链接](https://leetcode.cn/problems/implement-strstr/)
12+
[力扣题目链接](https://leetcode.cn/problems/find-the-index-of-the-first-occurrence-in-a-string/)
1313

1414
实现 strStr() 函数。
1515

@@ -699,7 +699,7 @@ class Solution(object):
699699
if haystack[i:i+n]==needle:
700700
return i
701701
return -1
702-
```
702+
```
703703
```python
704704
// 方法一
705705
class Solution:

‎problems/0031.下一个排列.md‎

Lines changed: 19 additions & 58 deletions
Original file line numberDiff line numberDiff line change
@@ -160,73 +160,34 @@ class Solution {
160160
```
161161
162162
## Python
163-
>直接使用sorted()不符合题意
163+
>直接使用sorted()会开辟新的空间并返回一个新的list,故补充一个原地反转函数
164164
```python
165165
class Solution:
166166
def nextPermutation(self, nums: List[int]) -> None:
167167
"""
168168
Do not return anything, modify nums in-place instead.
169169
"""
170-
for i in range(len(nums)-1, -1, -1):
171-
for j in range(len(nums)-1, i, -1):
170+
length = len(nums)
171+
for i in range(length - 1, -1, -1):
172+
for j in range(length - 1, i, -1):
172173
if nums[j] > nums[i]:
173174
nums[j], nums[i] = nums[i], nums[j]
174-
nums[i+1:len(nums)] = sorted(nums[i+1:len(nums)])
175-
return
176-
nums.sort()
177-
```
178-
>另一种思路
179-
```python
180-
class Solution:
181-
'''
182-
抛砖引玉:因题目要求"必须原地修改,只允许使用额外常数空间",python内置sorted函数以及数组切片+sort()无法使用。
183-
故选择另一种算法暂且提供一种python思路
184-
'''
185-
def nextPermutation(self, nums: List[int]) -> None:
186-
"""
187-
Do not return anything, modify nums in-place instead.
188-
"""
189-
length = len(nums)
190-
for i in range(length-1, 0, -1):
191-
if nums[i-1] < nums[i]:
192-
for j in range(length-1, 0, -1):
193-
if nums[j] > nums[i-1]:
194-
nums[i-1], nums[j] = nums[j], nums[i-1]
195-
break
196-
self.reverse(nums, i, length-1)
197-
break
198-
if n == 1:
199-
# 若正常结束循环,则对原数组直接翻转
200-
self.reverse(nums, 0, length-1)
201-
202-
def reverse(self, nums: List[int], low: int, high: int) -> None:
203-
while low < high:
204-
nums[low], nums[high] = nums[high], nums[low]
205-
low += 1
206-
high -= 1
207-
```
208-
>上一版本简化版
209-
```python
210-
class Solution(object):
211-
def nextPermutation(self, nums: List[int]) -> None:
212-
n = len(nums)
213-
i = n-2
214-
while i >= 0 and nums[i] >= nums[i+1]:
215-
i -= 1
216-
217-
if i > -1: // i==-1,不存在下一个更大的排列
218-
j = n-1
219-
while j >= 0 and nums[j] <= nums[i]:
220-
j -= 1
221-
nums[i], nums[j] = nums[j], nums[i]
175+
self.reverse(nums, i + 1, length - 1)
176+
return
177+
self.reverse(nums, 0, length - 1)
222178
223-
start, end = i+1, n-1
224-
while start < end:
225-
nums[start], nums[end] = nums[end], nums[start]
226-
start += 1
227-
end -= 1
228-
229-
return nums
179+
def reverse(self, nums: List[int], left: int, right: int) -> None:
180+
while left < right:
181+
nums[left], nums[right] = nums[right], nums[left]
182+
left += 1
183+
right -= 1
184+
185+
"""
186+
265 / 265 个通过测试用例
187+
状态:通过
188+
执行用时: 36 ms
189+
内存消耗: 14.9 MB
190+
"""
230191
```
231192

232193
## Go

‎problems/0040.组合总和II.md‎

Lines changed: 36 additions & 31 deletions
Original file line numberDiff line numberDiff line change
@@ -258,40 +258,45 @@ public:
258258
**使用标记数组**
259259
```Java
260260
class Solution {
261-
List<List<Integer>> lists = new ArrayList<>();
262-
Deque<Integer> deque = new LinkedList<>();
263-
int sum = 0;
264-
265-
public List<List<Integer>> combinationSum2(int[] candidates, int target) {
266-
//为了将重复的数字都放到一起,所以先进行排序
267-
Arrays.sort(candidates);
268-
//加标志数组,用来辅助判断同层节点是否已经遍历
269-
boolean[] flag = new boolean[candidates.length];
270-
backTracking(candidates, target, 0, flag);
271-
return lists;
272-
}
261+
LinkedList<Integer> path = new LinkedList<>();
262+
List<List<Integer>> ans = new ArrayList<>();
263+
boolean[] used;
264+
int sum = 0;
273265

274-
public void backTracking(int[] arr, int target, int index, boolean[] flag) {
275-
if (sum == target) {
276-
lists.add(new ArrayList(deque));
277-
return;
278-
}
279-
for (int i = index; i < arr.length && arr[i] + sum <= target; i++) {
280-
//出现重复节点,同层的第一个节点已经被访问过,所以直接跳过
281-
if (i > 0 && arr[i] == arr[i - 1] && !flag[i - 1]) {
282-
continue;
283-
}
284-
flag[i] = true;
285-
sum += arr[i];
286-
deque.push(arr[i]);
287-
//每个节点仅能选择一次,所以从下一位开始
288-
backTracking(arr, target, i + 1, flag);
289-
int temp = deque.pop();
290-
flag[i] = false;
291-
sum -= temp;
292-
}
266+
public List<List<Integer>> combinationSum2(int[] candidates, int target) {
267+
used = new boolean[candidates.length];
268+
// 加标志数组,用来辅助判断同层节点是否已经遍历
269+
Arrays.fill(used, false);
270+
// 为了将重复的数字都放到一起,所以先进行排序
271+
Arrays.sort(candidates);
272+
backTracking(candidates, target, 0);
273+
return ans;
274+
}
275+
276+
private void backTracking(int[] candidates, int target, int startIndex) {
277+
if (sum == target) {
278+
ans.add(new ArrayList(path));
279+
}
280+
for (int i = startIndex; i < candidates.length; i++) {
281+
if (sum + candidates[i] > target) {
282+
break;
283+
}
284+
// 出现重复节点,同层的第一个节点已经被访问过,所以直接跳过
285+
if (i > 0 && candidates[i] == candidates[i - 1] && !used[i - 1]) {
286+
continue;
287+
}
288+
used[i] = true;
289+
sum += candidates[i];
290+
path.add(candidates[i]);
291+
// 每个节点仅能选择一次,所以从下一位开始
292+
backTracking(candidates, target, i + 1);
293+
used[i] = false;
294+
sum -= candidates[i];
295+
path.removeLast();
293296
}
297+
}
294298
}
299+
295300
```
296301
**不使用标记数组**
297302
```Java

‎problems/0046.全排列.md‎

Lines changed: 1 addition & 1 deletion
Original file line numberDiff line numberDiff line change
@@ -331,7 +331,7 @@ var permute = function(nums) {
331331

332332
```
333333

334-
## TypeScript
334+
### TypeScript
335335

336336
```typescript
337337
function permute(nums: number[]): number[][] {

‎problems/0063.不同路径II.md‎

Lines changed: 44 additions & 14 deletions
Original file line numberDiff line numberDiff line change
@@ -210,24 +210,54 @@ public:
210210
```java
211211
class Solution {
212212
public int uniquePathsWithObstacles(int[][] obstacleGrid) {
213-
int n = obstacleGrid.length, m = obstacleGrid[0].length;
214-
int[][] dp = new int[n][m];
215-
216-
for (int i = 0; i < m; i++) {
217-
if (obstacleGrid[0][i] == 1) break; //一旦遇到障碍,后续都到不了
218-
dp[0][i] = 1;
213+
int m = obstacleGrid.length;
214+
int n = obstacleGrid[0].length;
215+
int[][] dp = new int[m][n];
216+
217+
//如果在起点或终点出现了障碍,直接返回0
218+
if (obstacleGrid[m - 1][n - 1] == 1 || obstacleGrid[0][0] == 1) {
219+
return 0;
219220
}
220-
for (int i =0; i < n; i++) {
221-
if (obstacleGrid[i][0] == 1) break; ////一旦遇到障碍,后续都到不了
222-
dp[i][0] = 1;
221+
222+
for (int i =0; i < m &&obstacleGrid[i][0] == 0; i++) {
223+
dp[i][0] = 1;
223224
}
224-
for (int i = 1; i < n; i++) {
225-
for (int j = 1; j < m; j++) {
226-
if (obstacleGrid[i][j] == 1) continue;
227-
dp[i][j] = dp[i - 1][j] + dp[i][j - 1];
225+
for (int j = 0; j < n && obstacleGrid[0][j] == 0; j++) {
226+
dp[0][j] = 1;
227+
}
228+
229+
for (int i = 1; i < m; i++) {
230+
for (int j = 1; j < n; j++) {
231+
dp[i][j] = (obstacleGrid[i][j] == 0) ? dp[i - 1][j] + dp[i][j - 1] : 0;
232+
}
233+
}
234+
return dp[m - 1][n - 1];
235+
}
236+
}
237+
```
238+
239+
```java
240+
// 空间优化版本
241+
class Solution {
242+
public int uniquePathsWithObstacles(int[][] obstacleGrid) {
243+
int m = obstacleGrid.length;
244+
int n = obstacleGrid[0].length;
245+
int[] dp = new int[n];
246+
247+
for (int j = 0; j < n && obstacleGrid[0][j] == 0; j++) {
248+
dp[j] = 1;
249+
}
250+
251+
for (int i = 1; i < m; i++) {
252+
for (int j = 0; j < n; j++) {
253+
if (obstacleGrid[i][j] == 1) {
254+
dp[j] = 0;
255+
} else if (j != 0) {
256+
dp[j] += dp[j - 1];
257+
}
228258
}
229259
}
230-
return dp[n - 1][m -1];
260+
return dp[n - 1];
231261
}
232262
}
233263
```

‎problems/0070.爬楼梯.md‎

Lines changed: 14 additions & 23 deletions
Original file line numberDiff line numberDiff line change
@@ -213,22 +213,6 @@ public:
213213
214214
215215
### Java
216-
```Java
217-
class Solution {
218-
public int climbStairs(int n) {
219-
// 跟斐波那契数列一样
220-
if(n <= 2) return n;
221-
int a = 1, b = 2, sum = 0;
222-
223-
for(int i = 3; i <= n; i++){
224-
sum = a + b;
225-
a = b;
226-
b = sum;
227-
}
228-
return b;
229-
}
230-
}
231-
```
232216
233217
```java
234218
// 常规方式
@@ -241,15 +225,22 @@ public int climbStairs(int n) {
241225
}
242226
return dp[n];
243227
}
228+
```
229+
230+
```Java
244231
// 用变量记录代替数组
245-
public int climbStairs(int n) {
246-
int a = 0, b = 1, c = 0; // 默认需要1次
247-
for (int i = 1; i <= n; i++) {
248-
c = a + b; // f(i - 1) + f(n - 2)
249-
a = b; // 记录上一轮的值
250-
b = c; // 向后步进1个数
232+
class Solution {
233+
public int climbStairs(int n) {
234+
if(n <= 2) return n;
235+
int a = 1, b = 2, sum = 0;
236+
237+
for(int i = 3; i <= n; i++){
238+
sum = a + b; // f(i - 1) + f(i - 2)
239+
a = b; // 记录f(i - 1),即下一轮的f(i - 2)
240+
b = sum; // 记录f(i),即下一轮的f(i - 1)
241+
}
242+
return b;
251243
}
252-
return c;
253244
}
254245
```
255246

‎problems/0078.子集.md‎

Lines changed: 0 additions & 4 deletions
Original file line numberDiff line numberDiff line change
@@ -180,10 +180,6 @@ class Solution {
180180
List<List<Integer>> result = new ArrayList<>();// 存放符合条件结果的集合
181181
LinkedList<Integer> path = new LinkedList<>();// 用来存放符合条件结果
182182
public List<List<Integer>> subsets(int[] nums) {
183-
if (nums.length == 0){
184-
result.add(new ArrayList<>());
185-
return result;
186-
}
187183
subsetsHelper(nums, 0);
188184
return result;
189185
}

‎problems/0112.路径总和.md‎

Lines changed: 2 additions & 2 deletions
Original file line numberDiff line numberDiff line change
@@ -499,7 +499,7 @@ class solution:
499499
if not cur_node.left and not cur_node.right:
500500
if remain == 0:
501501
result.append(path[:])
502-
return
502+
return
503503

504504
if cur_node.left:
505505
path.append(cur_node.left.val)
@@ -508,7 +508,7 @@ class solution:
508508

509509
if cur_node.right:
510510
path.append(cur_node.right.val)
511-
traversal(cur_node.right, remain-cur_node.left.val)
511+
traversal(cur_node.right, remain-cur_node.right.val)
512512
path.pop()
513513

514514
result, path = [], []

‎problems/0135.分发糖果.md‎

Lines changed: 7 additions & 10 deletions
Original file line numberDiff line numberDiff line change
@@ -135,25 +135,22 @@ class Solution {
135135
2、起点下标 ratings.length - 2 从右往左, 只要左边 比 右边 大,此时 左边的糖果应该 取本身的糖果数(符合比它左边大) 和 右边糖果数 + 1 二者的最大值,这样才符合 它比它左边的大,也比它右边大
136136
*/
137137
public int candy(int[] ratings) {
138-
int[] candyVec = new int[ratings.length];
138+
int len = ratings.length;
139+
int[] candyVec = new int[len];
139140
candyVec[0] = 1;
140-
for (int i = 1; i < ratings.length; i++) {
141-
if (ratings[i] > ratings[i - 1]) {
142-
candyVec[i] = candyVec[i - 1] + 1;
143-
} else {
144-
candyVec[i] = 1;
145-
}
141+
for (int i = 1; i < len; i++) {
142+
candyVec[i] = (ratings[i] > ratings[i - 1]) ? candyVec[i - 1] + 1 : 1;
146143
}
147144
148-
for (int i = ratings.length - 2; i >= 0; i--) {
145+
for (int i = len - 2; i >= 0; i--) {
149146
if (ratings[i] > ratings[i + 1]) {
150147
candyVec[i] = Math.max(candyVec[i], candyVec[i + 1] + 1);
151148
}
152149
}
153150
154151
int ans = 0;
155-
for (int s : candyVec) {
156-
ans += s;
152+
for (int num : candyVec) {
153+
ans += num;
157154
}
158155
return ans;
159156
}

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